Javascript：为什么前缀运算符使用模数而不是后缀运算符？(Javascript: Why does prefix operator work with modulus but not post

Javascript：为什么前缀运算符使用模数而不是后缀运算符？(Javascript: Why does prefix operator work with modulus but not postfix operator?)

我试图使一个函数递增，直到它达到3然后从零开始（因此，调用三次它将记录0然后1然后2当使用%运算符与前后修复操作符时，我有困惑结果。

这是我的两个功能：

var i, j = 0, 0 function run () { console.log(i); i = i++ % 3; } // Called three times logs 0, 0, 0

和

function newRun () { console.log(j); j = ++j % 3; } // Called three times it logs 0, 1, 2

为什么前缀运算符工作而后缀不工作（即在第一个函数中为什么i从不递增？

I am trying to make a function that increments until it reaches 3 and then starts back from zero (so, called three times it would log 0 then 1 then 2. When using the % operator with the pre and post fix operators, I have confusing results.

Here are my two functions:

var i, j = 0, 0 function run () { console.log(i); i = i++ % 3; } // Called three times logs 0, 0, 0

And

function newRun () { console.log(j); j = ++j % 3; } // Called three times it logs 0, 1, 2

Why does the prefix operator work and the postfix does not (i.e. in the first function why is i never incremented?

最满意答案

这与模运算符没有任何关系。 甚至

i = i++;

不起作用 - 它取一个值，递增它，然后用最初获取的值覆盖它。 另请参阅循环中i ++和++ i的区别？ 他们如何工作。

你可能想写

i = (i + 1) % 3;

This doesn't have anything to do with the modulo operator. Even

i = i++;

doesn't work - it takes a value, increments it, and then overwrites it with the initially taken value. See also Difference between i++ and ++i in a loop? for how they work.

You probably want to write

i = (i + 1) % 3;